Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $k = \dfrac{5y^2 - 45y}{5y^3 - 35y^2 - 90y} \times \dfrac{2y^2 - 8y - 24}{y - 5} $
Solution: First factor out any common factors. $k = \dfrac{5y(y - 9)}{5y(y^2 - 7y - 18)} \times \dfrac{2(y^2 - 4y - 12)}{y - 5} $ Then factor the quadratic expressions. $k = \dfrac {5y(y - 9)} {5y(y + 2)(y - 9)} \times \dfrac {2(y + 2)(y - 6)} {y - 5} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {5y(y - 9) \times 2(y + 2)(y - 6) } { 5y(y + 2)(y - 9) \times (y - 5)} $ $k = \dfrac {10y(y + 2)(y - 6)(y - 9)} {5y(y + 2)(y - 9)(y - 5)} $ Notice that $(y + 2)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {10y\cancel{(y + 2)}(y - 6)(y - 9)} {5y\cancel{(y + 2)}(y - 9)(y - 5)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $k = \dfrac {10y\cancel{(y + 2)}(y - 6)\cancel{(y - 9)}} {5y\cancel{(y + 2)}\cancel{(y - 9)}(y - 5)} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $k = \dfrac {10y(y - 6)} {5y(y - 5)} $ $ k = \dfrac{2(y - 6)}{y - 5}; y \neq -2; y \neq 9 $